If $\phi$ is the work function of photosensitive material in eV and light of wavelength of numerical value $\lambda=\dfrac{hc}{e}$ metre, is incident on it with energy above its threshold value at an instant then the maximum kinetic energy of the photoelectron ejected by it at that instant (Take h - Plank's constant, c - velocity of light in free space) is (in SI units):

A $e+2\phi$

B $2e-\phi$

C $e-\phi$

D $e+\phi$

The answer is C.

Solution

$(KE)_{max}=\dfrac{hc}{\lambda}-\phi_0$. Given $\lambda=\dfrac{hc}{e}$, so $\dfrac{hc}{\lambda}=e$. Hence $(KE)_{max}=e-\phi$.

Concept Photoelectric effect - Einstein's photoelectric equation

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